Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.5 - Solving Equations by the Zero-Factor Property - 5.5 Exercises - Page 355: 20


$z=\left\{ -3,3 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Express the given equation, $ 9z^2-81=0 ,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ The factored form of the equation above is \begin{array}{l}\require{cancel} 9(z^2-9)=0 \\\\ 9(z+3)(z-3)=0 \\\\ \dfrac{9(z+3)(z-3)}{9}=\dfrac{0}{9} \\\\ (z+3)(z-3)=0 .\end{array} Equating each factor to zero (Zero-Factor Property), then \begin{array}{l}\require{cancel} z+3=0 \text{ OR } z-3=0 .\end{array} Using the properties of equality to solve each of the equation above results to \begin{array}{l}\require{cancel} z+3=0 \\\\ z=-3 \\\\\text{ OR }\\\\ z-3=0 \\\\ z=3 .\end{array} Hence, the solutions are $ z=\left\{ -3,3 \right\} .$ $\bf{\text{Supplementary Solution/s:}}$ Factoring the $GCF= 9 ,$ the expression, $ 9z^2-81 ,$ above is equivalent to \begin{array}{l}\require{cancel} 9 \left( \dfrac{9z^2}{9}-\dfrac{81}{9} \right) \\\\= 9 \left( z^2-9 \right) .\end{array} Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} 9(z+3)(z-3) .\end{array}
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